Algebra

Question 1
If x3 + y3 = 9 and x + y = 3, then the value of x4+y4 is,
A
21
B
0
C
17
D
25
Algebra    
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Question 1 Explanation: 
x3+y3 = (x + y) × (x2 − xy + y2)

Putting given values of x3+y3 and (x + y)
9 = 3 × ((x+y)2 − 3xy)
  = 3 × (9 − 3xy) 
  = 27 − 9xy

9xy = 18
xy = 2

x4 + y4 = (x2 + y2)2 - 2x2y2
   = (x2 + y2)2 - 2*4 
                                  [Putting value of xy]
   = ((x + y)2 - 2xy)2 - 2*4 
                            [Putting values of (x+y) and xy]
   = (9 - 4)2 - 2*4 
   = 17
Question 2
If x+1/2x = 2, find the value of 8x3+1/x3
A
40
B
20
C
28
D
35
Algebra    
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Question 3
For any real number x the maximum value of 4−6x−x2 is at x=,
A
4
B
6
C
-3
D
3
Algebra    
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Question 3 Explanation: 
Differentiate and equate to 0 6+2x =0 x=−3
Question 4
If 5√x +12√x =13√x then value of x is,
A
2
B
1
C
3
D
4
Algebra    
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Question 4 Explanation: 
say x=1
17!=13

for, x=2 and 3 also not possible
x=4
52+122=132
169=169
Question 5
1_q
A
0
B
-2
C
2
D
4
Algebra    
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Question 5 Explanation: 
1_q_sol
Question 6
2_q
A
0
B
6
C
-4
D
4
Algebra    
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Question 6 Explanation: 
2_q_sol
Question 7
3_q
A
8
B
2
C
0
D
6
Algebra    
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Question 7 Explanation: 
3_qsol
Question 8
4_q
A
7/4
B
9/4
C
5/4
D
3/4
Algebra    
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Question 8 Explanation: 
4_q_sol
Question 9
5_q
A
6
B
2
C
3
D
0
Algebra    
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Question 9 Explanation: 
Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac
(ab + bc + ca)/[(a+b+c)(abc)]
as, ab+bc+ca=0
= 0
Question 10
Following instructions are to be used throughout the quiz: Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read both the statements and Give answer (a) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question. Give answer (b) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question. Give answer (c) if the data either in Statement I or in Statement II alone are sufficient to answer the question. Give answer (d) if the data even in both Statements I and II together are not sufficient to answer the question. Give answer (e) if the data in both Statements I and II together are necessary to answer the question. If x,y are integers, then (x2 + y2)1/2 is an integer? I) x2 + y2 is an integer II) x2 - 3y2 = 0
A
A
B
B
C
C
D
D
E
E
Algebra    Data Sufficiency    
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Question 11
4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number?
A
630
B
63
C
315
D
105
Algebra    
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Question 11 Explanation: 
Let the number be x. Then, 4/15 of 5x/7 = 4/9 of 2x/5 + 8 4/15 * 5x/7 = 4/9 * 2x/5 + 8 4x/21 = 8x/45 + 8 x/21 = 2x/45 + 1 x/21 = (2x + 45)/45 45x = 21(2x + 45) 45x = 42x + 45*21 3x = 45*21 x = 15*21 = 315.
Question 12
If 2x/(1 + 1/(1 + x/(1 - x))) = 1, then find the value of x.
A
1/3
B
2/3
C
1/2
D
3/2
Algebra    
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Question 12 Explanation: 
We have 2x/(1 + 1/(1 + x/(1-x))) = 1 2x/(1 + 1/((1-x+x)/(1-x))) = 1 2x/(1 + 1/(1/(1-x))) = 1 2x/(1 + 1-x) = 1 2x/(2-x) = 1 2x = 2-x 3x = 2 x = 2/3.
Question 13
Find the value of (1-1/3)(1-1/4)(1-1/5)...(1-1/100).
A
1/10
B
1/50
C
1/20
D
1/100
Algebra    
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Question 13 Explanation: 
We have (1-1/3)(1-1/4)(1-1/5)...(1-1/100) = (2/3)(3/4)(4/5)...(99/100) = 2/100 = 1/50.
Question 14
A zoo has some peacocks and some horses. If the number of heads be 48 and the number of feet be 140, then the number of peacocks will be:
A
23
B
25
C
26
D
27
Algebra    
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Question 14 Explanation: 
Let there be x peacocks and y horses. Then, x + y = 48 (equation 1) and 2x + 4y = 140 x + 2y = 70 Substituting the value of x from equation 1, we get 48 -y + 2y = 70 y = 22 Then, x = 48 - 22 = 26.
Question 15
A total of 324 coins of 20 paise and 25 paise make a sum of Rs 71. The number of 25-paise coins is:
A
104
B
120
C
128
D
124
Algebra    
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Question 15 Explanation: 
Let the number of 20-paise coins be x and the number of 25-paise coins be y. Then, x + y = 324 (equation 1) And, x/5 + y/4 = 71 4x + 5y = 71*20 (Multiplying by 20 on both sides) Substituting the value of x from equation 1, we get 4(324-y) + 5y = 71*20 1296 - 4y + 5y = 1420 y = 124.
Question 16
mock
A
1
B
2
C
3
D
0
Algebra    GATE 2017 Mock    
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Question 16 Explanation: 
mock_14
Correction: Instead of R3 -> R3 + 3R it should be R3 -> R3 + 3R1 while applying the first set of rules.
There are 16 questions to complete.

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